How do you multiply #(4x + 5)^2#?

Question

Isaiah Anthony · Accepted Answer

#16x^2+40x+25# #(4x+5)^2=(4x+5)(4x+5)# #"each term in the second factor is multiplied by "# #"each term in the first factor"# #rArr(color(red)(4x+5))(4x+5)# #=color(red)(4x)(4x+5)color(red)(+5)(4x+5)# #"distribute each product"# #=16x^2+20x+20x+25# #=16x^2+40x+25larrcolor(blue)"collect like terms"#

Eli Assouline · Answer

#16x^2+40x+25#

Use the FOIL method, or the shortcut for squaring binomials.

For any #x# and #y#, #(x+y)^2=x^2+2xy+y^2#

Our #x# is #4x# and our #y# is #5#.

Therefore:

#(4x+5)^2=(4x)^2+2(4x*5)+(5)^2#

#(4x+5)^2=16x^2+40x+25#

Alternatively, you could multiply 'each by each,' using the FOIL method.

Eva Adamson · Answer

See a solution process below:

We can use this rule for quadratics: #(color(red)(x) + color(blue)(y))^2 = (color(red)(x) + color(blue)(y))(color(red)(x) + color(blue)(y)) = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2# Let: #color(red)(x) = 4x# #color(blue)(y) = 5# Substituting gives: #(color(red)(4x) + color(blue)(5))^2 =># #(color(red)(4x) + color(blue)(5))(color(red)(4x) + color(blue)(5)) =># #16x^2 + 40x + 25#

Eva Adamson · Answer

#=16x^2 + 40x + 25#

#(4x + 5)^2#

As you can see, each users are solving it differently and this is what makes Math funny and unique. There are different steps, strategies you can use to solve a problem.

Well, I'm gonna use my favorite method which is Distributive

So let's start:

#=(4x)(4x)+(4x)(5)+(5)(4x)+(5)(5)#

#=16x^2 + 20x + 20x + 25#

#=16x^2 + 40x + 25#

Savannah Anderson · Answer

undefined To multiply (4x + 5)^2, you use the formula for squaring a binomial, which is (a + b)^2 = a^2 + 2ab + b^2.

In this case, a = 4x and b = 5. So,

(4x + 5)^2 = (4x)^2 + 2(4x)(5) + (5)^2
            = 16x^2 + 40x + 25.